Write an expression for the work done when a force is acting on an object in the direction of its displacement

NCERT Solutions for Class 9 Science Chapter 11: Work and Energy NCERT Solutions Class 9 Science Chapter 11 – CBSE Free PDF Download * According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10. NCERT Solutions for Class 9 Science Chapter 11 Work and Energy help you lay a good foundation for your CBSE exam preparation. Students who refer to NCERT Solutions for Class 9 Science Chapter 11: Work and Energy help students to practise and gain more confidence in their preparations. All the Previous Next  Access Answers of Science NCERT Class 9 Chapter 11: Work and Energy (All intext and exercise questions solved) Exercise-11.1 Page: 148 1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force. Let us take it that the force acts on the object through the displacement. What is the work done in this case? Solution: When a force F acts on an object to move it in its direction through a distance S, work is done. The work on the body is done by force. Work done = Force × Displacement W = F × S Where, F = 7 N S = 8 m So, work done, W = 7 × 8 W = 56 Nm W = 56 J Exercise-11.2 Page: 149 1. When do we say that work is done? Solution: Work is completed whenever the given conditions are satisfied: (i) A force acts on the body. (ii) There’s a displacement of the body by applying force in or opposite to the direction of the force. 2. Write an expression for the work done when a force is acting on an object in the dire...

Definition: Work Done on a Body by a Force The work done on a body by a force is dependent on the force that acts on the body and the distance that the body moves in the direction of that force according to the formula 𝑊 = 𝐹 ⋅ 𝑑 ( 𝜃 ) , c o s where 𝐹 is the magnitude of the force, 𝑑 is the magnitude of displacement of the body while the force acts on it, and 𝜃 is the angle between the directions of 𝐹 and 𝑑. An alternative way of representing the work done on a body by a force is to represent the force and the displacement as vectors rather than as the magnitudes of vectors. The product of two vectors ⃑ 𝑎 and ⃑ 𝑏 can be the dot product of the vectors, which is defined as follows. The work done by a force with magnitude 𝐹 over a displacement with magnitude 𝑑 is equal to ⃑ 𝐹 ⋅ ⃑ 𝑑 ⃑ 𝐹 ⋅ ⃑ 𝑑 = ‖ ‖ ⃑ 𝐹 ‖ ‖ ‖ ‖ ⃑ 𝑑 ‖ ‖ ( 𝜃 ) . c o s A graphical representation of ⃑ 𝐹 and ⃑ 𝑑 demonstrates that the product of the magnitude of ⃑ 𝐹 and the magnitude of the component of ⃑ 𝑑 in the direction of ⃑ 𝐹 is equal to ‖ ‖ ⃑ 𝐹 ‖ ‖ ‖ ‖ ⃑ 𝑑 ‖ ‖ ( 𝜃 ) c o s. The product is given by ⃑ 𝐹 ⋅ ⃑ 𝑑 = ( 4 , − 3 ) ⋅ ( 3 , 4 ) ⃑ 𝐹 ⋅ ⃑ 𝑑 = ( 4 × 3 ) + ( − 3 × 4 ) = 1 2 − 1 2 = 0 . The scalar product of two perpendicular vectors is zero. There is no force acting on the body in the direction of the displacement, and so the force does no work on the body. Let us look at an example of using vector notation to determine the work done by a force. Example 1: Calculating the Work Done by a Force Acting on a Particle...

The normal force and the object's weight are in static equilibrium (they are balanced forces), the applied force, F, is an unbalanced force and will result in the object being accelerated across the top of the table's surface in the same direction as the force. This acceleration will change the object's velocity and subsequently its kinetic energy. We say that this applied force is doing work on the object. The amount of work done by F is directly proportional to the distance through which the force is applied as it pulls the object across the table's surface. This statement tells us that when an external force does work on an object it will change the object's kinetic energy; that is, it will cause the object to either gain or lose speed. When more than one force is acting on an object, all forces that are either parallel or antiparallel to the direction the object moves will do work. If the object's velocity remains constant, that just means that the work done by opposing forces (for example, a forward applied force, F, and an opposing force, friction) are equal. Note that if F and s are perpendicular to each other no work is done on the object. In our example of the block being dragged across the table, neither the normal force nor the weight would do any work on the block since they act at right angles to the direction of the block's motion. Another example would be when a satellite is being held in circular orbit by the force of gravity. Note that since the satellite'...

Learning Objectives By the end of this section, you will be able to: • Apply the work-energy theorem to find information about the motion of a particle, given the forces acting on it • Use the work-energy theorem to find information about the forces acting on a particle, given information about its motion We have discussed how to find the work done on a particle by the forces that act on it, but how is that work manifested in the motion of the particle? According to Newton’s second law of motion, the sum of all the forces acting on a particle, or the net force, determines the rate of change in the momentum of the particle, or its motion. Therefore, we should consider the work done by all the forces acting on a particle, or the net work, to see what effect it has on the particle’s motion. Let’s start by looking at the net work done on a particle as it moves over an infinitesimal displacement, which is the dot product of the net force and the displacement: d W net = F → net · d r → . d W net = F → net · d r → . Newton’s second law tells us that F → net = m ( d v → / d t ) , F → net = m ( d v → / d t ) , so d W net = m ( d v → / d t ) · d r → . d W net = m ( d v → / d t ) · d r → . For the mathematical functions describing the motion of a physical particle, we can rearrange the differentials dt, etc., as algebraic quantities in this expression, that is, d W net = m ( d v → d t ) · d r → = m d v → · ( d r → d t ) = m v → · d v → , d W net = m ( d v → d t ) · d r → = m d v → · ...

Learning Objectives By the end of this section, you will be able to: • Represent the work done by any force • Evaluate the work done for various forces In physics, work is done on an object when energy is transferred to the object. In other words, work is done when a force acts on something that undergoes a displacement from one position to another. Forces can vary as a function of position, and displacements can be along various paths between two points. We first define the increment of work dW done by a force [latex] \overset [/latex] as the dot product of these two vectors: We choose to express the dot product in terms of the magnitudes of the vectors and the cosine of the angle between them, because the meaning of the dot product for work can be put into words more directly in terms of magnitudes and angles. We could equally well have expressed the dot product in terms of the various components introduced in x– and y-components in Cartesian coordinates, or the r– and [latex] \phi [/latex]-components in polar coordinates; in three dimensions, it was just x-, y-, and z-components. Which choice is more convenient depends on the situation. In words, you can express Recall that the magnitude of a force times the cosine of the angle the force makes with a given direction is the component of the force in the given direction. The components of a vector can be positive, negative, or zero, depending on whether the angle between the vector and the component-direction is between [...

Work, Energy and Power Work, Energy and Power are fundamental concepts of Physics. Work is said to be done when a force (push or pull) applied to an object causes a displacement of the object. We define the capacity to do the work as energy. Power is the work done per unit of time. This article discusses work, energy and power in detail. Table of Contents • • • • • • • • • • • • • • • • What is Work? For work to be done, a force must be exerted and there must be motion or displacement in the direction of the force. The work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction. Hence, work is a scalar quantity. \(\begin \) Where W is the work done, F is the force, d is the displacement, θ is the angle between force and displacement and F cosθ is the component of force in the direction of displacement. We understand from the work equation that if there is no displacement, there is no work done, irrespective of how large the force is. To summarize, we can say that no work is done if: • the displacement is zero • the force is zero • the force and displacement are mutually perpendicular to each other. Can the work done be negative? Watch the video and find out! Example of Work An object is horizontally dragged across the surface by a 100 N force acting parallel to the surface. Find out the amount of work done by the force in moving the object through ...